Calculate $\lim_{n \rightarrow \infty } \frac{n (1- na_n)}{\log n} $

limitsreal-analysis

Given the recursive sequence $\{a_n\}$ defined by setting $0 < a_1 < 1, \; a_{n+1} = a_n(1-a_n) , \; n \ge 1 $

Calculate : $$\lim_{n \rightarrow \infty } \frac{n (1- na_n)}{\log n} $$

My attempts : $$\lim_{n \rightarrow \infty } \frac{n (1- n a_n)}{\log n} =\lim_{n \rightarrow \infty }\frac {n \left (\frac{1}{n a_n} -1 \right) n a_n} {\log n}= \lim_{n \rightarrow \infty } \frac {\frac{1}{a_n} – n}{ \log n}$$

Now I am not able to proceed further.

Please help me.

Thank You.

Best Answer

Note. The answer below may be excessive compared to what OP is asking.

For a quick answer, read the definition of $(x_n)$, jump directly to the proof of proposition, and then read only the first 3 steps.

Let $a_1 \in (0, 1)$ and define $x_n = 1/a_n$. Then $x_n$ solves the following recurrence relation

$$ x_{n+1} = x_n + 1 + \frac{1}{x_n} + \frac{1}{x_n(x_n - 1)}. \tag{1}$$

Using this we progressively reveal the asymptotic behavior of $(x_n)$. More precisely, our goal is to prove the following statement.

Proposition. Let $(x_n)$ be defined by $\text{(1)}$, i.e. $x_1 > 1$ and $x_{n+1} = f(x_n)$ for $f(x) = \frac{x^2}{x-1}$. Then there exists a function function $C : (1, \infty) \to \mathbb{R}$ such that

$$ x_n = n + \log n + C(x_1) + \mathcal{O}\left(\frac{\log n}{n}\right) \quad \text{as} \quad n\to\infty. $$

Here, the implicit constant of the asymptotic notation may depend on $x_1$. Moreover, $C$ solves the functional equation $C(f(x)) = C(x) + 1$.

We defer the proof to the end and analyze the asymptotic behavior of OP's limit first. Plugging the asymptotic expansion of $x_n$, we find that

$$ r_n := \frac{n(1-n a_n)}{\log n} = \frac{n(x_n - n)}{x_n \log n} = 1 + \frac{C(x_1)}{\log n} + \mathcal{O}\left(\frac{\log n}{n}\right). $$

This tells that, not only that $r_n \to 1$ as $n\to\infty$, but also that the convergence is extremely slow due to the term $C/\log n$.

For instance, $f^{\circ 94}(2) \approx 100.37$ tells that $C(100) \approx C(2) + 94$. Indeed, a numerical simulation using $n = 10^6$ shows that

\begin{align*} x_1 = 2 &\quad \Rightarrow \quad (r_n - 1)\log n \approx 0.767795, \\ x_1 = 100 &\quad \Rightarrow \quad (r_n - 1)\log n \approx 94.3883, \end{align*}

which loosely matches the prediction above.

Proof of Proposition.

Step 1. Since $x_{n+1} \geq x_n + 1$, it follows that $x_n \geq n + \mathcal{O}(1)$. In particular, $x_n \to \infty$ as $n\to\infty$.

Step 2. Since $x_{n+1} - x_n \to 1$, we have $\frac{x_n}{n} \to 1$ by Stolz-Cesaro theorem.

Step 3. Using the previous step, we find that

$$ \frac{x_{n+1} - x_n - 1}{\log(n+1) - \log n} = \frac{1}{(x_n - 1)\log\left(1+\frac{1}{n}\right)} \xrightarrow[n\to\infty]{} 1 $$

So, again by Stolz-Cesaro theorem, we have $x_{n+1} = n + (1+o(1))\log n$. This is already enough to conclude that OP's limit is $1$.

Step 4. By the previous step, we find that $ x_{n+1} - x_n = 1 + \frac{1}{n} + \mathcal{O}\left(\frac{\log n}{n^2}\right)$. Using this, define $C$ by the following convergent series

$$ C(x_1) = x_1 - 1 + \sum_{n=1}^{\infty} \underbrace{ \left( x_{n+1} - x_n - 1 - \log\left(1+\frac{1}{n}\right) \right) }_{=\mathcal{O}(\log n/n^2)}. $$

Splitting the sum for $n < N$ and $n \geq N$ and using the estimate $\sum_{n\geq N}\frac{\log n}{n^2} = \mathcal{O}\left(\frac{\log n}{n}\right)$,

$$ C(x_1) = x_N - N - \log N + \mathcal{O}\left(\frac{\log N}{N}\right), $$

which confirms the first assertion of the proposition.

Once this is established, then the second assertion easily follows by interpreting $x_{n+1}$ as the $n$-th term of the sequence that solves $\text{(1)}$ with the initial value $f(x_1)$. Hence comparing both

$$ x_{n+1} = n+1 + \log(n+1) + C(x_1) + o(1) $$

and

$$ x_{n+1} = n + \log n + C(f(x_1)) + o(1) $$

the second assertion follows. ////

Related Question